Q.Two sitar strings $A$ and $B$ playing the note 'Ga' are slightly out of tune and produce beats of frequency $6\ \text{Hz}$. The tension in the string $A$ is slightly reduced and the beat frequency is found to reduce to $3\ \text{Hz}$. If the original frequency of $A$ is $324\ \text{Hz}$, what is the frequency of $B$?

Showing the 12 most recent of 14 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Two identical strings each of length 0.750 m are each tuned exactly to 440 Hz. The tension in one of the strings is then increased by 1.0 %. If they are now struck, then the beat frequency between the fundamentals of the two strings is nearly (A) 1 (B) 2 (C) 3 (D) No beats

   ›Reveal solutionSolution

   A small 1% increase in tension raises the string's frequency by about half that percentage (since $f\propto\sqrt{T}$), producing a beat frequency of roughly 2 Hz between the two strings.

   Concept and Intuition

   The fundamental frequency of a vibrating string is $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, so for fixed length and linear mass density, $f \propto \sqrt{T}$. A small fractional change in tension $T$ produces roughly half that fractional change in $f$ (by the binomial approximation $\sqrt{1+x}\approx 1+x/2$ for small $x$). When two originally identical strings' tensions differ slightly, they produce two slightly different frequencies, and superposing their sounds produces beats at the difference frequency.

   Step-by-Step Solution

   1. Both strings start identical: $f_1 = f_2 = 440\ \text{Hz}$ (same $L$, $\mu$, $T$).
   2. Tension in one string is increased by $1\% = 0.01$: $T' = 1.01\,T$.
   3. New frequency: $f' = f\sqrt{T'/T} = f\sqrt{1.01}$.
   4. Using the small-x approximation, $\sqrt{1.01} \approx 1 + \dfrac{0.01}{2} = 1.005$.
   5. $f' \approx 440 \times 1.005 = 442.2\ \text{Hz}$. …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Two vibrating tuning forks produce progressive waves given by $y_1 = 2\sin(500\pi t - ax)$ and $y_2 = 2\sin(506\pi t - bx)$. Number of beats produced per minute is (A) 360 (B) 180 (C) 60 (D) 3

   ›Reveal solutionSolution

   Read off each wave's frequency from its angular frequency ($\omega=2\pi f$), subtract to get the beat frequency (3 Hz), then convert to beats per minute — 180.

   Concept and Intuition

   When two waves of slightly different frequency superpose, the resulting amplitude modulation ("beats") occurs at a rate equal to the difference of the two frequencies. Reading $\omega$ straight from the argument of each sine function gives each wave's frequency directly.

   Step-by-Step Solution

   1. $y_1 = 2\sin(500\pi t - ax) \implies \omega_1 = 500\pi\text{ rad/s} \implies f_1 = \dfrac{\omega_1}{2\pi} = 250$ Hz.
   2. $y_2 = 2\sin(506\pi t - bx) \implies \omega_2 = 506\pi\text{ rad/s} \implies f_2 = \dfrac{\omega_2}{2\pi} = 253$ Hz.
   3. Beat frequency $= |f_2-f_1| = 253-250 = 3$ Hz $=3$ beats per second.
   4. Beats per minute $= 3 \times 60 = 180$. …

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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Ten tuning forks are arranged in increasing order of frequency. Any two nearest tuning forks produce 4 beats per second. The highest frequency of tuning fork is twice that of the lowest possible, then the highest and lowest frequencies, in Hz, are respectively (A) 80 and 40 (B) 100 and 60 (C) 72 and 32 (D) 72 and 36

   ›Reveal solutionSolution

   This tests setting up an arithmetic progression of frequencies from the "beats between adjacent forks" condition, then using the given ratio between highest and lowest to solve for the actual values.

   Concept and Intuition

   When ten tuning forks are arranged in increasing order of frequency and any two adjacent forks produce a fixed beat frequency (here 4 Hz), their frequencies form an arithmetic progression with common difference equal to that beat frequency. With $n=10$ forks there are $n-1 = 9$ gaps between the lowest and the highest fork.

   Step-by-Step Solution

   1. Let the lowest frequency be $f$ (Hz). Successive fork frequencies: $f,\ f+4,\ f+8,\ \dots,\ f+9\times4$ (10 terms total, so 9 steps of 4 Hz).
   2. Highest frequency $= f + 36$.
   3. Given: highest $= 2\times$ lowest, i.e. $f + 36 = 2f$.
   4. Solve: $36 = 2f - f = f \Rightarrow f = 36\ \text{Hz}$.
   5. Highest frequency $= f + 36 = 36+36 = 72\ \text{Hz}$. …

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4. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Two tuning forks when sounded together produced 6 beats per second. The frequency of one tuning fork is 256 $\mathrm{Hz}$. When small piece of wax is attached to the second tuning fork, they produced 2 beats per second. The original frequency of second tuning fork is (A) 250 $\mathrm{Hz}$ (B) 254 $\mathrm{Hz}$ (C) 258 $\mathrm{Hz}$ (D) 262 $\mathrm{Hz}$

   ›Reveal solutionSolution

   Beat frequency fixes $f_2$ to 250 Hz or 262 Hz; the effect of wax (frequency decreases) picks 262 Hz as the only consistent choice.

   Concept and Intuition

   Beat frequency is $|f_1 - f_2|$, so it never tells you which fork is higher — that ambiguity is resolved by a physical perturbation like adding wax, filing, or loading with tape, all of which only ever reduce a tuning fork's frequency (added mass/wax lowers the natural frequency).

   Step-by-Step Solution

   1. Given $f_1 = 256$ Hz and beat frequency $= 6$ Hz: $f_2 = 256 \pm 6$, i.e. $f_2 = 250$ Hz or $f_2 = 262$ Hz.
   2. Wax is attached to fork 2, and its frequency can only decrease.
   3. Case $f_2 = 262$ Hz: after wax, $f_2' < 262$. New beats $= 2 \Rightarrow f_2' = 254$ or $258$ Hz. Since $f_2'$ decreased from 262, and moving down towards 256 Hz reduces the beat count, $f_2' = 258$ Hz (still above 256, beats $=2$) is consistent with a small decrease. …

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5. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Two waves of wavelengths 50 cm and 51 cm produced 12 beats per second. The velocity of sound is (A) $306\,\mathrm{ms^{-1}}$ (B) $331\,\mathrm{ms^{-1}}$ (C) $340\,\mathrm{ms^{-1}}$ (D) $360\,\mathrm{ms^{-1}}$

   ›Reveal solutionSolution

   Beat frequency is the difference of the two individual frequencies; solving $\Delta f=v(1/\lambda_1-1/\lambda_2)=12$ gives $v\approx306\,\mathrm{ms^{-1}}$.

   Concept and Intuition

   When two waves of slightly different frequency (or wavelength) superpose, the resulting beat frequency is the magnitude of the difference between the two individual frequencies: $f_{beat}=|f_1-f_2|$. Using $f=v/\lambda$ lets us relate the beat frequency directly to the speed of sound.

   Step-by-Step Solution

   1. $f_1=\dfrac{v}{\lambda_1}=\dfrac{v}{0.50}$, $f_2=\dfrac{v}{\lambda_2}=\dfrac{v}{0.51}$ (in metres).
   2. Beat frequency: $f_{beat}=f_1-f_2=v\left(\dfrac1{0.50}-\dfrac1{0.51}\right)$.
   3. $\dfrac1{0.50}=2$, $\dfrac1{0.51}\approx1.960784$.
   4. $f_{beat}=v(2-1.960784)=v(0.039216)$. …

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6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.Two sound waves of wavelengths 99 cm and 100 cm produce 10 beats in a time of t seconds. If the speed of sound in air is $330\ \mathrm{m\,s^{-1}}$, then the value of t in seconds is (A) 12 (B) 9 (C) 6 (D) 3

   ›Reveal solutionSolution

   Two close wavelengths produce a beat frequency computed from their corresponding frequencies; 10 beats occur in 3 seconds.

   Concept and Intuition

   Beats arise when two sound waves of slightly different frequency interfere, producing a periodic rise and fall in loudness at the difference frequency $|f_1-f_2|$. Frequency is related to wavelength via $f=v/\lambda$, so a small difference in wavelength (99 cm vs 100 cm) translates into a small but nonzero beat frequency.

   Step-by-Step Solution

   1. $f_1 = v/\lambda_1 = 330/0.99\ \mathrm{Hz}$, $f_2 = v/\lambda_2 = 330/1.00 = 330\ \mathrm{Hz}$.
   2. Beat frequency: $\Delta f = f_1-f_2 = v\left(\dfrac{1}{\lambda_1}-\dfrac{1}{\lambda_2}\right) = 330\left(\dfrac{1}{0.99}-1\right) = 330\times\dfrac{1-0.99}{0.99} = 330\times\dfrac{0.01}{0.99} = \dfrac{3.3}{0.99} = \dfrac{10}{3}\ \mathrm{Hz}$. …

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7. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.When a stretched string is vibrated simultaneously with a 440 Hz tuning fork, a beat frequency of 5 Hz is produced. If the experiment is repeated with a tuning fork of 437 Hz, the beat frequency produced is 8 Hz. The frequency of the string is (A) 445 Hz (B) 435 Hz (C) 429 Hz (D) 448 Hz

   ›Reveal solutionSolution

   Beats only tell you the magnitude of the frequency difference, giving two candidate string frequencies per tuning fork; the value common to both trials, 445 Hz, is the answer.

   Concept and Intuition

   When two frequencies close to each other are sounded together, they produce beats at a rate equal to the magnitude of their difference: $f_{beat}=|f_1-f_2|$. This is ambiguous on its own (the unknown could be above or below the known frequency), so a second trial with a different reference frequency is used to break the tie — only one candidate value will be consistent with both observations.

   Step-by-Step Solution

   1. With the 440 Hz fork, beat frequency 5 Hz $\Rightarrow$ string frequency $f = 440\pm5 = 445$ or $435$ Hz.
   2. With the 437 Hz fork, beat frequency 8 Hz $\Rightarrow$ string frequency $f = 437\pm8 = 445$ or $429$ Hz. …

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8. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If two progressive sound waves represented by $y_1 = 3\sin 250\pi t$ and $y_2 = 2\sin 260\pi t$ (where displacement is in metre and time is in second) superimpose, then the time interval between two successive maximum intensities is (A) 0.1 s (B) 0.4 s (C) 0.5 s (D) 0.2 s

   ›Reveal solutionSolution

   This tests the beats phenomenon: two close frequencies superpose to give a periodic intensity maximum, recurring at the reciprocal of the beat frequency — here $0.2\ \text{s}$.

   Concept and Intuition

   When two waves of slightly different frequencies $f_1$ and $f_2$ superimpose, their resultant amplitude waxes and wanes at the beat frequency $f_{beat}=|f_1-f_2|$. Since intensity depends on amplitude squared, a maximum of loudness (maximum intensity) occurs once every beat period $T_{beat}=1/f_{beat}$.

   Step-by-Step Solution

   1. Identify angular frequencies from $y_1=3\sin(250\pi t)$ and $y_2=2\sin(260\pi t)$: $\omega_1=250\pi$, $\omega_2=260\pi$.
   2. Convert to ordinary frequency using $f=\omega/2\pi$: $f_1 = 125\ \text{Hz}$, $f_2=130\ \text{Hz}$.
   3. Beat frequency: $f_{beat}=|f_2-f_1| = 5\ \text{Hz}$. …

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9. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.On a sonometer, the lengths of two wires are in the ratio 35 : 34, diameters are in the ratio 4 : 1, densities are in the ratio 1 : 2 and the tensions in the string are in the ratio 8 : 1. If the note of higher pitch has frequency of 350 Hz then find the frequency of the beats produced when sounded together (A) $20\ \mathrm{Hz}$ (B) $7\ \mathrm{Hz}$ (C) $5\ \mathrm{Hz}$ (D) $10\ \mathrm{Hz}$

   ›Reveal solutionSolution

   Using $f\propto \dfrac{1}{Ld}\sqrt{T/\rho}$ for each wire and the given ratios, the two frequencies are 340 Hz and 350 Hz, giving 10 Hz beats.

   Concept and Intuition

   A stretched wire's fundamental frequency is $f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ where $\mu=\rho A=\rho\dfrac{\pi d^2}{4}$ is mass per unit length. So $f\propto\dfrac{1}{Ld}\sqrt{\dfrac{T}{\rho}}$ (constants cancel in a ratio). Comparing the two wires via their given ratios of L, d, ρ, T lets us find the frequency ratio directly.

   Step-by-Step Solution

   1. Let wire-1: $L_1=35,\ d_1=4,\ \rho_1=1,\ T_1=8$; wire-2: $L_2=34,\ d_2=1,\ \rho_2=2,\ T_2=1$.
   2. $\dfrac{f_1}{f_2}=\dfrac{L_2 d_2}{L_1 d_1}\sqrt{\dfrac{T_1\rho_2}{T_2\rho_1}}=\dfrac{34\times1}{35\times4}\sqrt{\dfrac{8\times2}{1\times1}}=\dfrac{34}{140}\sqrt{16}=\dfrac{34}{140}\times4=\dfrac{136}{140}=\dfrac{34}{35}$. …

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10. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.Two stretched strings A and B when vibrated together produce 4 beats per second. If the tension applied to the string A increased, the number of beats produced per second is increased to 7. If the frequency of string B is 480 Hz initially, the frequency of string A is (A) 473 Hz (B) 476 Hz (C) 484 Hz (D) 487 Hz

    ›Reveal solutionSolution

    The tension-increase test resolves the two possible frequencies of string A; since beats grow when tension (and hence $f_A$) increases, A's frequency must already exceed B's.

    Concept and Intuition

    Beat frequency only tells you $|f_A-f_B|$, not which is larger — that ambiguity is exactly what the "increase tension and watch the beats" test resolves, since raising tension always raises a string's fundamental frequency.

    Step-by-Step Solution

    1. $|f_A-480|=4 \Rightarrow f_A=476$ Hz or $f_A=484$ Hz.
    2. Increasing the tension in A increases $f_A$.
    3. If $f_A=476$ Hz (below 480), increasing $f_A$ moves it toward 480, which would first decrease the beat frequency — contradicting the given increase to 7 Hz. …

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11. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.Beats are produced by frequencies $\upsilon_1$ and $\upsilon_2$ ($\upsilon_1 > \upsilon_2$). The duration of time between two successive minima is (A) $\dfrac{1}{\upsilon_1 + \upsilon_2}$ (B) $\dfrac{2}{\upsilon_1 - \upsilon_2}$ (C) $\dfrac{2}{\upsilon_1 + \upsilon_2}$ (D) $\dfrac{1}{\upsilon_1 - \upsilon_2}$

    ›Reveal solutionSolution

    The time between two successive beat-minima equals one full beat period, $\dfrac{1}{\upsilon_1-\upsilon_2}$.

    Concept and Intuition

    When two waves of close frequencies superpose, the resultant amplitude is modulated by an envelope $|\cos(\pi\Delta\upsilon\, t)|$, where $\Delta\upsilon=\upsilon_1-\upsilon_2$. This envelope reaches zero (silence/minimum) once per beat period — so the time between two consecutive minima is exactly one beat period, not half of it.

    Step-by-Step Solution

    1. Beat frequency $= \upsilon_1-\upsilon_2$ (given $\upsilon_1>\upsilon_2$).
    2. Resultant intensity envelope $\propto \cos^2(\pi\Delta\upsilon\,t)$ (or equivalently $|\cos(\pi\Delta\upsilon t)|$ for amplitude), which is zero when $\pi\Delta\upsilon\, t = \pi/2, 3\pi/2, 5\pi/2,\dots$
    3. Consecutive zero times: $t_n = \dfrac{(2n-1)}{2\Delta\upsilon}$. The spacing between consecutive $t_n$ is $\dfrac{1}{\Delta\upsilon}$. …

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12. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.Two open organ pipes of lengths 50 cm and 51 cm are totally immersed in a medium. They are found to give 40 beats in 10 s when each is sounding at its fundamental note. The speed of sound in this medium is (A) 275 $\mathrm{ms^{-1}}$ (B) 310 $\mathrm{ms^{-1}}$ (C) 258 $\mathrm{ms^{-1}}$ (D) 204 $\mathrm{ms^{-1}}$

    ›Reveal solutionSolution

    Setting the difference of the two open-pipe fundamental frequencies equal to the observed 4 Hz beat rate gives the speed of sound in the medium as 204 m/s.

    Concept and Intuition

    An open organ pipe of length $L$ has fundamental frequency $f=\dfrac{v}{2L}$ (both ends being antinodes, so the pipe holds half a wavelength). Two pipes of slightly different lengths therefore have slightly different fundamentals, and their difference is the beat frequency heard when both sound together.

    Step-by-Step Solution

    1. $f_1 = \dfrac{v}{2(0.50)} = \dfrac{v}{1.00}$ (for $L_1=50\,\mathrm{cm}=0.50\,\mathrm m$).
    2. $f_2 = \dfrac{v}{2(0.51)} = \dfrac{v}{1.02}$ (for $L_2=51\,\mathrm{cm}=0.51\,\mathrm m$).
    3. Beats per second $=\dfrac{40\text{ beats}}{10\text{ s}}=4\,\mathrm{Hz}$.
    4. $f_1 - f_2 = v\left(\dfrac1{1.00}-\dfrac1{1.02}\right) = v\cdot\dfrac{1.02-1.00}{1.00\times1.02} = v\cdot\dfrac{0.02}{1.02}$. …

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